When sodium (Na) reacts with chlorine (Cl), they combine to form sodium chloride (NaCl). To determine how much sodium chloride will form when 31.3 g of sodium and 35.45 g of chlorine are combined, we need to consider the molar masses and stoichiometry of the reaction.
1. Convert the masses of sodium and chlorine to moles:
- The molar mass of sodium (Na) is approximately 22.99 g/mol. Therefore, 31.3 g of sodium is equal to 31.3 g / 22.99 g/mol = 1.36 mol of sodium.
- The molar mass of chlorine (Cl) is approximately 35.45 g/mol. Therefore, 35.45 g of chlorine is equal to 35.45 g / 35.45 g/mol = 1 mol of chlorine.
2. Determine the limiting reactant:
To determine the limiting reactant, we compare the moles of each reactant. The reactant that has fewer moles is the limiting reactant because it will be completely consumed in the reaction. In this case, both sodium and chlorine have a 1:1 stoichiometric ratio, meaning that 1 mol of sodium reacts with 1 mol of chlorine to form 1 mol of sodium chloride.
Since we have 1.36 mol of sodium and only 1 mol of chlorine, chlorine is the limiting reactant.
3. Calculate the moles and mass of sodium chloride formed:
Since the stoichiometric ratio between sodium and sodium chloride is 1:1, the number of moles of sodium chloride formed will be equal to the number of moles of sodium. Therefore, 1.36 mol of sodium chloride will be formed.
To calculate the mass of sodium chloride, we multiply the number of moles by the molar mass of sodium chloride, which is approximately 58.44 g/mol. Therefore, the mass of sodium chloride formed is 1.36 mol * 58.44 g/mol = 79.39 g.
Therefore, when 31.3 g of sodium and 35.45 g of chlorine are combined, approximately 79.39 g of sodium chloride will form.